JEE Main 202330 Jan, Morning shiftMathematicsEasySequence and Series179× since 2002If an=−24n2−16n+15{a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}an=4n2−16n+15−2, then a1+a2 + .... + a25{a_1} + {a_2}\, + \,....\, + \,{a_{25}}a1+a2+....+a25 is equal to :AOption A. 51144{{51} \over {144}}14451BOption B. 49138{{49} \over {138}}13849COption C. 50141{{50} \over {141}}14150DOption D. 52147{{52} \over {147}}14752CheckSkip · show solution
